Khí cầu có dung tích
Giải thích
\(\frac{{pV}}{T} = nR \Rightarrow \frac{{{{1,5.1,013.10}^5}.328}}{{27 + 273}} = n.8,31 \Rightarrow n \approx 19992\;{\rm{mol}}\)
\(m = nM = 19992.2 = 39983\;{\rm{g}} \approx 40\;{\rm{kg}}\)
\(\frac{{pV}}{T} = nR \Rightarrow \frac{{{{1,5.1,013.10}^5}.328}}{{27 + 273}} = n.8,31 \Rightarrow n \approx 19992\;{\rm{mol}}\)
\(m = nM = 19992.2 = 39983\;{\rm{g}} \approx 40\;{\rm{kg}}\)